Download Part 3 Answers Only for Questions, Exercises, and Problems in The

Instructor’s Manual to Accompany Cracolice/Peters 3/e
Part 3
Answers Only for Questions, Exercises,
and Problems in The Active Learning Workbook
CHAPTER 2
 denotes problems assignable in OWL.
2.  Macroscopic: c. Microscopic: a. Particulate: b.
4. Chemists use models, real or imaginary or both, to represent the invisible particles that make up matter.
6. Ice consists of water molecules that vibrate or shake in fixed positions relative to one another. The ice
has a fixed shape. When the ice melts, the molecules remain together but move freely among each other at
the bottom of the container that holds them, assuming the shape of the container. As a gas, the molecules
separate, moving freely throughout a closed container or escaping into the atmosphere from an open
container.
8. The particles in a solid occupy fixed positions relative to each other and cannot be poured, but different
pieces of solids can move relative to each other. The slogan emphasizes that this brand of table salt has
solid pieces small enough to move freely relative to one another, but not so small that they stick together in
humid weather.
10. Liquid volume is usually slightly greater than solid; gas volume is very much larger than liquid.
12.  Chemical: a. Physical: b, c.
14.  Chemical: a, b. Physical: c.
16. The change is chemical because the reactant diatomic molecules are destroyed, changing to monatomic
atoms.
18. If both products are the same pure substance, they must be identical; thus, purchasing the new, halfpriced product is the better choice.
20. The photograph alone does not provide enough information to answer the question. You probably
know, however, that dishwashing liquid is a mixture.
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22. All are mixtures.
24. (a) is a mixture because different substances are visible. (b) could be a pure substance in two different
states, but it is probably a mixture. (c) could be either a pure substance or a mixture because it may be one
kind of matter or two or more types of matter with similar appearances.
26. Yes, the terms homogeneous and heterogeneous refer to the macroscopic appearance of a sample. A
container filled with ice and liquid water is heterogeneous in appearance but is also pure, as long as in both
phases the water is pure.
28.  Homogeneous: a, c. Heterogeneous: b.
30. The cylinder appears the same throughout, so it is a homogeneous substance.
32. Your sketch should include two or more different types of particles mixed together (a mixture), evenly
distributed in the same phase of matter (so that it will appear homogeneous).
34. Pick out table tennis balls or ball bearings based on size and appearance; use magnetic properties of
steel to pick up ball bearings with a magnet; use the lower density of balls and higher density of steel to
float the balls in water. These methods are based on physical properties.
36. Place the mixture in water. Salt will dissolve and sand will not. Filter out the sand. Evaporate the water
to obtain pure salt. Dissolving and recrystallizing the salt is a physical change because the salt did not
undergo a chemical change (you can taste the salt dissolved in the water, so it is still salt).
38.  Elements: b, c. Compound: a.
40.  Elements: a, c. Compound: b.
42. Elements: a, d, e. Compound: c. Neither element nor compound (not pure): b.
44. (a) Elements: 2, 3. Compounds: 1, 4, 5. (b) In general, if there are two or more words in the name, the
substance is a compound. However, many compounds are known by one-word common names, and oneword names for many carbon-containing compounds are assembled from prefixes, suffixes, and special
names for recurring groups.
46. Only a compound with one taste can be decomposed into a gas and a solid having a different taste.
48.
G, L, S
P, M
Hom, Het
E, C
Limestone (calcium carbonate)
S
P
Hom
C
Lead
S
P
Hom
E
L&S
M
Het
G
P
Hom
Freshly squeezed orange juice
Oxygen
E
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Butter in the refrigerator
S
M
Hom
50. Attraction: a, c. Repulsion: b.
52.  Products: a, d. Reactants: b, c. Energy is released. The reaction is exothermic.
54. Product element: H2. Reactant compound: H2O.
56.  Endothermic: c. Exothermic: a, b.
58. Kinetic energy increases.
60. The masses would be the same. Products of a chemical reaction have the same total mass as reactants.
62. Potential energy of the water above the dam changes into kinetic energy as it falls into the turbine. The
turbine drives a generator in which kinetic energy is changed into the electrical energy that produces the
heat and light energy given off by the bulb.
66. Almost everything you use is either a result of or involves human-made chemical change in some way.
67. Almost everything you use is a mixture. Sugar is a rare example of a pure substance that you may have
used today.
68. Air, certainly, and perhaps tap water, and/or some glass or plastic products.
82. Your illustration will resemble the phases in Figure 2.5. According to kinetic molecular theory,
particles move faster at higher temperatures, so particles are more likely to separate from one another.
CHAPTER 3
 denotes problems assignable in OWL
2.  (a) 7.03 × 104 (b) 2.31 × 10–2 (c) 1.54 × 10–4 (d) 5.04 × 103
4.  (a) 0.0232 (b) 92,700 (c) 2540 (d) 0.000896
6.  (a) 9.30 × 10–12 (b) 1.58 (c) 1.40 × 1013
8.  (a) 9.90 × 1011 (b) 5.03 (c) 1.51 × 10–5
10.  (a) 8.61 × 10–4 (b) 2.46 × 109 (c) 9.48 × 104
12.  (a) 8.63 × 105 (b) 1.89 × 10–9 (c) 6.82 × 10–8
14.  615 mi
16.  623 days
18.  864 eggs
20.  445 nickels
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22. 2,678,400 s
24. The mass of the rock is the same wherever it is. The weight of the rock is the same whether it is in the
lake or on the beach—or anywhere in the gravitational field on the earth’s surface. In the lake the upward
buoyancy force cancels some of the downward weight force. That makes the rock seem lighter.
26. The gram (the SI unit of mass is the kilogram, 1000 g)
28. Kilounit is a general term that refers to 1000 units of any measurement unit. Kilogram is a specific
example—1000 grams.
30. 100 cm = 1 m
32. An nm is a nanometer. It is a very short distance, one billionth of a meter, or about 4/100,000,000 of an
inch.
34.  (a) 1.53 × 107 mg; (b) 0.0805 kg, 8.05 × 104 mg; (c) 5.85 × 10–5 kg; 0.0585 g
36.  (a) 0.0904 m; 9.04 cm; (b) 1.19 × 104 mm; 1.19 × 103 cm; (c) 536 mm; 0.536 m
38.  (a) 0.0908 L; 90.8 cm3; (b) 1.69 × 104 mL; 1.69 × 104 cm3; (c) 65.4 mL; 0.0654 L
40. (a) 1.94 × 108 g; (b) 5.66 × 10–9 m; (c) 4.81 × 105 cm
42. (a) 5, (b) 2, (c) 3, (d) uncertain—3 or 4, (e) 3, (f) 5, (g) 3, (h) 5
44. (a) 52.2 mL; (b) 18.0 g; (c) 78.5 mg; (d) 2.36 × 107 µm; (e) 0.00420 kg
46. 254 g
48. 6.88 g
50. 62.8 g; 231 g
52. 1.03 g/mL
54. 5.10 gal; 0.436 L
56.  91.7 oz
58. 0.313 oz; 8.88 g
60. 418 kg
62. 3.4 kg
64.  0.776 in.
66. 3.784 × 103 km
68. 0.5 mm
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70.  2.40 gal
72.
Celsius
Fahrenheit
Kelvin
4
40
277
317
603
590
–13
9
260
–44
–47
229
440
824
713
–192
–314
81
74.  T°C = 23°C; TK = 296 K
76.  T°F = 1.2 × 102°F; TK = 322 K
78.  T°C = 181°C; T°F = 358°F
80.  (a) d ∝ t; (b) d = s × t, s = 61.6 mi/hr; (c) 10.7 hr
82.  (a) V ∝ T, V = b × T; (b) b = 0.330 L/K; (c) 124 L
84.  (a) m ∝ V, m = D × V; (b) 185 g; (c) 21.8 cm3
86.  D = 11.5 g/mL; Table 3.4 lists the density of lead as 11.4 g/cm3
88. 7.87 g/cm3
90.  46.2 g
92.  1.73 mL
100. 1.6 m
102. 30.0 kg
104. 280°C
106. 8.3 lb
CHAPTER 4
 denotes problems assignable in OWL
2. Kinetic refers to the fact that gas particles are in motion. Kinetic energy is energy of motion.
4. Pressure results from the large numbers of particle collisions with the container walls.
6. Because gas molecules are widely spaced, air is compressible, which makes for a soft and comfortable
ride. The low density of air contributes little to the mass of an automobile. The constant motion of the
molecules makes the gas fill the tire, exerting pressure uniformly in all directions. There is no loss of
pressure because there is no loss of kinetic energy in molecular collisions in a gas.
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8. This is evidence that gas particles are moving.
10. Because gas molecules are more widely spaced than liquid molecules, the gas is less dense and
therefore rises through the liquid.
12. Gas particles are widely separated compared to the same number of particles close together in the liquid
state.
14. Particles move in straight lines until they hit something—eventually the walls of the container, thereby
filling it.
16. Quantity, volume, pressure, and temperature.
18. Barometer operation is explained in Figure 4.6.
20.
atm
1.84
0.946
0.959
0.984
psi
27.0
13.9
14.1
14.5
in. Hg
55.1
28.3
28.7
29.4
cm Hg
1.40 × 102
71.9
72.9
74.8
mm Hg
1.40 × 10
3
719
729
748
torr
1.40 × 103
719
729
748
Pa
1.86 × 105
9.59 × 104
9.72 × 104
9.97 × 104
kPa
186
95.9
97.2
99.7
bar
1.86
0.959
0.972
0.997
22. 752 torr – 284 torr = 468 torr
24. Absolute zero is the lowest temperature. It is a temperature that can only be approached but never
achieved. At absolute zero, particles have the minimum possible kinetic energy.
26.  263 K; 200 K
28.  311 K; 2467°C
30.  98°C; 2730°C
32.  10.3 L
34.  232°C
36. Reducing the volume increases the pressure and breaks the balloon.
38.  0.322 atm
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40.  942 mL
42.  308 torr
44.  156°C
46. Standard temperature and pressure for gases, 0°C and 1 atm.
48.  12.8 L
50.  8.80 L
52. 304 torr
54. –5°C
58. When gas particles are pushed close to each other, the intermolecular attractions become significant.
The molecules are no longer independent. Also, the volume of the molecules effectively reduces the
volume of the space available for the molecules to occupy. The ideal gas model is violated, so the gas does
not behave ideally.
60. 4.20 × 102 torr
62. 4.60 atm
64. 414 psi gauge
66. 1.26 m3
CHAPTER 5
 denotes problems assignable in OWL
2. See the summary in Section 5.1.
4. In a chemical reaction, the atoms in the reactant compounds are rearranged to form the product
compounds. Since atoms are not destroyed or created, the total mass must be the same before and after the
reaction.
6. They are the same because atoms cannot be created or destroyed.
8. The Law of Multiple Proportions for this case states that the same mass of chlorine, 10.0 g, will combine
with masses of mercury in a ratio of simple whole numbers. 56.6 ÷ 28.3 = 2, so the law is confirmed by
these data.
10.  a and d.
12. Most of the volume of a gold atom (any atom) is empty space.
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14. See the discussion in Section 5.3, including the summary box and the captions to Figures 5.3 and 5.4.
16. The Rutherford experiment showed that almost all of the mass of an atom is concentrated in an
extremely dense nucleus and that most of the atom’s volume is empty space occupied by electrons having
very small mass. When the massive proton and neutron were found, it was assumed they made up that
dense nucleus.
18. The number of protons is the same as the number of electrons. The number of neutrons is usually equal
to or greater than the number of protons or electrons.
20.  5 protons, 3 neutrons, and 5 electrons.
22. 
Name of
Nuclear
Atomic
Mass
Element
Symbol
Number
Number
Protons
Neutrons
Electrons
Lanthanum
138
57 La
57
138
57
81
57
Phosphorus
28
15 P
15
28
15
13
15
Lanthanum
!
139
57 La
57
139
57
82
57
Copper
!
65
29 Cu
29
65
29
36
29
Copper
!
63
29 Cu
29
63
29
34
29
Aluminum
!
25
13 Al
13
25
13
12
13
!
24. Exactly
! 1/12 the mass of a carbon-12 atom.
26. Boron occurs in nature as a mixture of atoms that have different masses.
28.  0.0009 × 137.9068 amu + 0.9991 × 138.9061 amu = 0.1 amu + 138.8 amu = 138.9 amu
30. 0.1978 × 10.0129 amu + (1.0000 – 0.1978) × 11.00931 amu = 1.981 amu + 8.832 amu = 10.813 amu.
The extra significant figure appears because adding two numerals in the units column produces a numeral
in the tens column.
32. 0.6909 × 62.9298 amu + 0.3091 64.9278 amu = 63.55 amu, Cu, copper
34. 0.3707 × 184.9530 amu + 0.6293 × 186.9560 amu = 186.2 amu, Re, rhenium
36. 0.6788 × 57.9353 amu + 0.2623 × 59.9332 amu + 0.0119 × 60.9310 amu + 0.0366 × 61.9283 amu +
0.0108 × 63.9280 amu = 58.73 amu, Ni, nickel
38.  (a) Cs, (b) S, (c) Mn, (d) Li
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40.  (a) Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn; (b) Na, Mg, Al; (c) O, S, Se, F, Cl, Br;
(d) Cs, Ba, Tl, Pb, Bi, Po
42.  (a) 14, (b) 6, (c) 29, (d) 12
44.  (a) 39.95 amu, (b) 69.72 amu, (c) 47.87 amu, (d) 24.31 amu
46. 
Table 5.5 Table of Elements
Name
Sodium
Lead
Aluminum
Iron
Fluorine
Boron
Argon
Silver
Carbon
Copper
Beryllium
Krypton
Chlorine
Hydrogen
Manganese
Chromium
Cobalt
Mercury
Atomic Number
11
82
13
26
9
5
18
47
6
29
4
36
17
1
25
24
27
80
Symbol of Element
Na
Pb
Al
Fe
F
B
Ar
Ag
C
Cu
Be
Kr
Cl
H
Mn
Cr
Co
Hg
51. The ratio of the masses of mercury is 402/201 = 2, a simple ratio of whole numbers.
53.  x × 137.9071 + (1 – x) × 138.9063 = 138.9; x = 0.0063
0.63% lanthanum-138 and 99.37% lanthanum-139
CHAPTER 6
 denotes problems assignable in OWL
2.  Diatomic molecules: hydrogen, H2; nitrogen, N2; oxygen, O2; fluorine, F2; chlorine, Cl2; bromine, Br2;
iodine, I2. Group 8A/18 elements: He, Ne, Ar, Kr, Xe, Rn.
4.  Ge, F2, Ar, Ga
6. Oxygen, calcium, barium, silver
8.  Cl2, Mg, N2, V
10.  Carbon tetrachloride, carbon tetrabromide, nitrogen monoxide, N2O, SO2, SF6
12.  Rb → Rb+ + e–
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14.  Calcium ion, phosphide ion, manganese(II) ion, telluride ion
16.  Ca2+, N3–, Mn2+, Li+
18. The element hydrogen is present in all acids discussed in this chapter.
20. The suffix -protic refers to protons, or hydrogen ions. A polyprotic acid is an acid that has two or more
ionizable hydrogens.
22.
Table 6.12
Acid Name
Hydrochloric
Nitric
Phosphoric
Hydrosulfuric
Nitrous
Iodic
Hydrotelluric
Hypochlorous
Iodous
Hydroselenic
Perchloric
Hydroiodic
Chlorous
Selenic
Bromous
Acid Formula
HCl
HNO3
H3PO4
H2S
HNO2
HIO3
H2Te
HClO
HIO2
H2Se
HClO4
HI
HClO2
H2SeO4
HBrO2
Ion Name
Chloride ion
Nitrate ion
Phosphate ion
Sulfide ion
Nitrite ion
Iodate ion
Telluride ion
Hypochlorite
Iodite
Selenide
Perchlorate
Iodide
Chlorite
Selenate
Bromite
Ion Formula
Cl–
NO3–
PO43–
S2–
NO2–
IO3–
Te2–
ClO–
IO2–
Se2–
ClO4–
I–
ClO2–
SeO42–
BrO2–
24.  H2SeO4 → H+ + HSeO4–; HSeO4– → H + + SeO42–
26.  H2PO4–, HSO4–, HTeO3–
28.  Hydrogen carbonate ion, hydrogen tellurate ion, hydrogen sulfite ion
30.  Acetic acid, gallium ion, permanganate ion
32.  C2H3O2–, Rb+, CrO42–
34.  Ba(NO2)2, NaHCO3, Al(NO3)3, CaCrO4
36.  Al(BrO2)3, K2Cr2O7, Ba(H2PO4)2, Rb2SeO4
38.  Co(OH)2, ZnCl2, Pb(NO3)2, CuNO3
40.  Lead(II) sulfide, silver hydroxide, cobalt(III) bromide
42.  Magnesium hydroxide, sodium carbonate, silver nitrite, cobalt(II) phosphate
44.  Barium bromate, potassium oxalate, aluminum hydrogen tellurate, potassium selenite
46.  KF • 2 H2O → KF + 2 H2O
48. Two; calcium chloride dihydrate
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50.  Barium perchlorate trihydrate; Ba(ClO4)2 • 3 H2O
52.  CaF2, NaBr, AlI3
54.  Sodium phosphate, calcium carbonate, aluminum acetate
56.  Sulfur trioxide, sodium oxide, (NH4)2SO4, N2O4
58. Hydrogen phosphate ion, copper(II) oxide, Na2C2O4, NH 3
60. HClO, CrBr2, potassium hydrogen carbonate, sodium dichromate
62. Cobalt(III) oxide, sodium sulfite, HgI2, Al(OH)3
64. Ca(H2PO4)2, KMnO4, ammonium iodate, selenic acid
66. Mercury(I) chloride, periodic acid, CoSO4, Pb(NO3)2
68. P4S7, BaO2, manganese(II) chloride, sodium chlorite
70. Potassium tellurate, zinc carbonate, CrCl2, HC2H3O2
72. BaCrO4 CaSO3, copper(I) chloride, silver nitrate
74. Sodium peroxide, nickel carbonate, FeO, H2S(aq) [The state designation (aq) distinguishes
hydrosulfuric acid from dihydrogen sulfide, H2S(g)]
76. Zn3P2, CsNO3, ammonium cyanide, disulfur decafluoride
78. Dinitrogen trioxide, lithium permanganate, In2Se3, Hg2(SCN)2
80. Cadmium chloride, nickel chlorate, CoPO4, Ca(IO4)2
CHAPTER 7
 denotes problems assignable in OWL
2.  (NH4)3PO4: 3 nitrogen atoms, 12 hydrogen atoms, 1 phosphorus atom, 4 oxygen atoms
4. Atomic mass is the mass of one atom; molecular mass is the mass of one molecule; formula mass is the
mass of one formula unit.
6. Atomic, molecular, and formula masses are expressed in atomic mass units, amu. An amu is exactly 1/12
the mass of a carbon-12 atom.
8.  a) NF3: 14.01 + 3(19.00) = 71.01 amu
b) BaCl2: 137.3 + 2(35.45) = 208.2 amu
c) Pb3(PO4)2: 3(207.2) + 2(30.97) + 8(16.00) = 811.5 amu
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10. The mole is the amount of any substance that contains the same number of units as the number of atoms
in exactly 12 grams of carbon-12. It is convenient to think of the mole as a number: 6.02 × 1023. The mole
is used to count the huge number of atoms and molecules in macroscopic samples.
12. Avogadro’s number, 6.02 × 1023
14.  (a) 7.53 × 1023 molecules; (b) 0.117 mol; (c) 0.131 mol; (d) 9.63 × 1023 molecules
16.  (a) 3.24 × 1024 H atoms; (b) 0.0231 mol O
18. Molar mass is the mass of one mole of molecules or formula units, expressed in grams per mole.
Molecular mass is the mass of a single molecule, expressed in amu.
20.  (a) 305.8 g/mol; (b) 60.09 g/mol; (c) 88.01 g/mol
22. (a) 5.97 mol Be; (b) 4.29 mol C3H4Cl4; (c) 0.0102 mol Ca(OH)2;
(d) 0.0333 mol CoBr3; (e) 0.0349 mol (NH4)2Cr2O7; (f) 0.127 mol Mg(ClO4)2
24. (a) 1.23 mol NaClO; (b) 4.32 mol Al(C2H3O2)3; (c) 0.00124 mol Hg2 Cl2
26. (a) 45.5 g NaHCO3; (b) 13.4 g AgNO3 ; (c) 1.36 × 103 g Na2HPO4; (d) 267 g Ca(BrO3)2;
(e) 132 g (NH4)2SO3
28. (a) 71.2 g MnO2; (b) 2.35 × 103 g Al(ClO3)3; (c) 114 g CrCl2
30. (a) 3.87 × 1023 Be(NO3)2 units; (b) 1.03 × 1023 Mn atoms’ (c) 9.50 × 1020 C3H7OH molecules
32. (a) 3.65 × 1022 I atoms; (b) 2.10 × 1021 C9H20 molecules; (c) 3.80 × 1023 MnCO3 units
34. (a) 30.8 g FeO; (b) 547 g F2: (c) 241 g Au
36. 5.01 × 1021 C atoms
38. 1.4 × 1025 C8H18 molecules
40. (a) 5.72 × 1022 F2 molecules: (b) 1.14 × 1023 F atoms: (c) 1.14 × 1023 F atoms: (d) 11.4 g F; (e) 22.8 g F2
42. a) Mg(NO3)2 is 16.39% Mg, 18.89% N, and 64.72% O.
b) Na3PO4 is 42.07% Na, 18.89% P, and 39.04% O.
c) CuCl2 is 47.27% Cu and 52.73% Cl.
d) Cr2(SO4)3 is 26.52% Cr, 24.53% S, and 48.95% O.
e) Ag2CO3 is 78.24% Ag, 4.354% C, and 17.40% O.
44. 6.12 g Br
46. 1.10 kg Mg
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48.  109 g C5H12O
50.  27.6 g NO
52.  22.6 g CO2
54. C2H6O and N2O5 are empirical formulas. The empirical formula of Na2O2 is NaO; the empirical
formula of C2H4O2 is CH2O.
56.

58.

60.

62.

64.

Element
Grams
Moles
1.475
Mole
Ratio
1.000
Formula
Ratio
1
B
15.94
F
C
H
O
S
F
S
O
F
S
F
84.06
39.12
8.772
52.11
21.96
78.04
31.42
31.35
37.23
25.24
74.76
4.424
3.257
8.702
3.257
0.6848
4.107
0.9797
1.959
1.959
0.7870
3.935
2.999
1.000
2.672
1.000
1.000
5.997
1.000
2.000
2.000
1.000
5.000
3
3
8
3
1
6
1
2
2
1
5
CHAPTER 8
Empirical
Formula
Molecular
Formula
BF3
C3H8O3
SF6
SO2F2
SF5
!
102.1
=1
102.07
SO2F2
254.1
=2
127.07
S2F10
!
 denotes problems assignable in OWL
2.  Particulate: A C2H4 molecule reacts with a hydrogen molecule to form a C2H6 molecule. Molar: One
mole of C2H4 reacts with one mole of hydrogen to form one mole of C2H6.
4. X2 + 3 Y2 → 2 XY3; The five X2 molecules require 15 or more Y2 molecules in the reactants box, and 10
XY3 molecules should be shown in the product box.
6. 2 Sb + 3 Cl2 → 2 SbCl3; reactants in box a; product in box c.
8. 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O; (a) Four butane molecules react with thirteen oxygen molecules to
form eight carbon dioxide molecules and ten water molecules; (b) 16 carbon atom models, 52 oxygen
atoms models, and 40 hydrogen atom models; (c) 8 carbon atom models, 26 oxygen atoms models, and 20
hydrogen atom models; (d) Four moles of butane reacts with thirteen moles of oxygen to form eight moles
of carbon dioxide and ten moles of water; (e) (2mol)(58 g/mol) + (13 mol)(32 g/mol) = 532 g = (8 mol)(44
g/mol) + (10 mol)(18 g/mol).
10.  N2(g) + 3 H2(g) → 2 NH3(g)
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12.  2 NO(g) + O2(g) → 2 NO2(g)
14.  P4(s) + 6 Cl2(g) → 4 PCl3(l)
16.  2 NaNO3(s) → 2 NaNO2(s) + O2(g)
18.  2 BrF3(l) → Br2(l) + 3 F2(g)
20.  Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g)
22.  Cu(s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq)
24.  2 CoI3(aq) + 3 Pb(NO3)2(aq) → 2 Co(NO3)3(aq) + 3 PbI2(s)
26.  Cr(NO3)3(aq) + 3 NaOH(aq) → Cr(OH)3(s) + 3 NaNO 3(aq)
28.  HClO4(aq) + NaOH(aq) → HOH(l) + NaClO4(aq)
30.  H2S(aq) + 2 NaOH(aq) → 2 HOH(l) + Na2S(aq)
32.  Ca(NO3)2(aq) + K2CO3(aq) → CaCO3(s) + 2 KNO3(aq)
34. 2 CH3CHO(l) + 5 O2(g) → 4 CO2(g) + 4 H2O(l)
36.  H2CO3(aq) → H2O(l) + CO2(g)
38. Ba(s) + 2 H2O(l) → Ba(OH)2(aq) + H2(g)
40.  Cl2(g) + 2 KBr(aq) → Br2(l) + 2 KCl(aq)
42. 2 AgNO3(aq) + Na2S(aq) → Ag2S(s) + 2 NaNO3(aq)
44.  4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)
46. 2 H2O2(l) → 2 H2O(l) + O2(g)
48.  HBr(aq) + KOH(aq) → HOH(l) + KBr(aq)
50. 2 F2(g) + O2(g) → 2 OF2(g)
52.  2 AgNO3(aq) + MgSO4(aq) → Ag2SO4(s) + Mg(NO3)2(aq)
54. 3 Mg(s) + N2(g) → Mg3N2(s)
56.  H2SO4(aq) + Zn(OH)2(aq) → 2 HOH(l) + ZnSO4(aq)
58. 2 Li(s) + MnCl2(aq) → 2 LiCl(aq) + Mn(s)
60.  Zn(s) + CuSO4(aq) → Cu(s) + ZnSO4(aq)
62. Zn(s) + H2O(g) → ZnO(s) + H2(g)
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64. BaO(s) + H2O(l) → Ba(OH)2(aq)
66. Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)
77. 2 LiOH(aq) + H2SO4(aq) → Li2SO4(aq) + 2 H2O(l)
79. H2C2O4(aq) + 2 NaOH(aq) → Na2C2O4(aq) + 2 HOH(l)
CHAPTER 9
 denotes problems assignable in OWL
2. When A dissolves in water, it forms ions. When B dissolves in water, it does not form ions.
4. When electricity passes through a wire, electrons “flow” through the wire. When electricity passes
through a solution, electrons are “carried” by charged ions that move freely. If a liquid can carry an
electrical current, it must be a solution containing ions.
6. Mg2+(aq), NO3–(aq); Fe3+(aq), Cl–(aq)
8.  HCN(aq); Na+(aq), NO3–(aq); H +(aq), ClO4–(aq)
10.  H+(aq), Cl–(aq); HCHO2(aq); K+(aq), I–(aq)
12.  Na+(aq), ClO4–(aq); H +(aq), I–(aq); HC2H3O2(aq)
14.  3 Ni2+(aq) + 2 PO43–(aq) → Ni3(PO4)2(s)
16.  Ni(s) + Cu2+(aq) → Cu(s) + Ni2+(aq)
18.  NO2–(aq) + H+(aq) → HNO2(aq)
20.  Sn(s) + Zn(NO3)2(aq) → NR
22.  Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s)
24.  Zn(s) + 2 H+(aq) → Zn2+(aq) + H2(g)
26.  2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(l)
28.  2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l)
30.  3 Co2+(aq) + 2 PO43–(aq) → Co3(PO4)2(s)
32.  Ba2+(aq) SO42–(aq) → BaSO4(s)
34.  Ni2+(aq) + 2 OH–(aq) → Ni(OH)2(s)
36.  NaNO3(aq) + Ba(OH)2(aq) → NR
38.  Pb2+(aq) + S2–(aq) → PbS(s); Cu2+(aq) + CO32–(aq) → CuCO3(s)
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40.  NO2–(aq) + H+(aq) → HNO2(aq)
42.  OH–(aq) + H+(aq) → H2O(l)
44.  HC2H3O2(aq) + OH–(aq) → HOH(l) + C2H3O2–(aq)
46.  2 H +(aq) + CaSO3(s) → H2O(l) + SO2(aq) + Ca2+(aq)
48.  NH4+(aq) + OH–(aq) → NH3(aq) + H2O(l)
50.  PO43–(aq) + 3 Ag+(aq) → Ag3PO4(s)
52.  Fe2(SO4)3(aq) + Co(NO3)2(aq) → NR
54.  Ni(s) + Pb2+(aq) → Ni2+(aq) + Pb(s)
56.  2 H +(aq) + CO32–(aq) → CO2(g) + H2O(l)
58.  OH–(aq) + H+(aq) → H2O(l)
60.  NaOH(aq) + KNO3(aq) → NR
62.  Cu(s) + Pb(NO3)2(aq) → NR
64.  CN–(aq) + H+(aq) → HCN(aq)
66.  OH–(aq) + NH4+(aq) → NH3(aq) + H2O(l)
68.  Ca2+(aq) + 2 OH–(aq) → Ca(OH)2(s)
70.  Pb(s) + 2 H+(aq) → Pb2+(aq) + H2(g)
72.  2 H +(aq) + Mn(OH)2(s) → Mn2+(aq) + 2 HOH(l)
74.  Ag+(aq) + Br–(aq) → AgBr(s)
76.  2 H +(aq) + CaSO3(s) → H2SO3(aq) + Ca2+(aq)
78.  HF(aq) + OH–(aq) → HOH(l) + F–(aq)
CHAPTER 10
 denotes problems assignable in OWL
2.  (a) 6 mol H2S; (b) 4 mol H2O: (c) 4 mol SO2
4.  2.54 mol CO2
6.  0.636 mol K2SO4
8. (a) 12.7 g C4H10; (b) 0.227 mol H2O; (c) 237 g CO2; (d) 101 g O2
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10.  79.1 g Cu(NO3)2
12.  237 g CCl4
14. 4.11 kg NaCl
16. (a) 3.90 kg HNO3; (b) 4.68 kg C7H5N3O6
18.  47.6 g NO
20. 377 g NH4HCO3
22. 7.10 × 102 g NaHCO3
24.  9.01 g NH3
26.  31.9 g HBr
28.  44.5 g CO
30.  51.9 g CaCO3
32.  88.0% yield
34.  5.51 g FeCl2 (act)
36.  3.67 g H2
38.  85.3% yield
40.  13.2 g NO2 (act)
42. Your product mixture box will show 6 CO2 molecules and 3 O2 molecules. CO is the limiting reactant.
The 6 CO molecules in the starting mixture can form a maximum of 6 CO2 molecules (2 mol CO/2 mol
CO2).
44. Hydrochloric acid is the limiting reactant in Flask 1. More than enough zinc was added to react with all
of the hydrochloric acid, and thus excess zinc remains. Flask 2 likely has stoichiometrically equivalent
amounts of zinc and hydrochloric acid. No zinc remains, but the balloon is inflated to the same volume as
in Flask 1, and Flask 1 had excess zinc, so the product hydrogen gas was at its maximum volume in both
Flask 1 and Flask 2. Zinc is the limiting reactant in Flask 3. All of the zinc is consumed, but the reaction
has not gone to the same volume of hydrogen as in Flasks 1 and 2, indicating that additional hydrochloric
acid is available to react.
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46.  CO2 is the limiting reactant. The yield is the smaller amount, 49.0 g K2CO3. 3.2 g KOH left.
48.  O2 is the limiting reactant. The yield is the smaller amount, 14.4 g CO2. 3.5 g CO left.
50. Na2CO3 is the limiting reactant, so 135 g Na2CO3 is not enough to neutralize 188 g HNO3. 55.9 grams
of CO2 is released.
52.  (a) 1.20 × 102 cal; 0.504 kJ; (b) 803 J; 0.803 kJ; (c) 423 J; 101 cal
54.  1.39 × 103 kJ; 1.39 × 106 J
56. 935 kJ
58.  N2(g) + 2 O2(g) + 66.4 kJ → 2 NO2(g)
N2(g) + 2 O2(g) → 2 NO2(g)
60.  NH3(g) +
NH3(g) +
5
4
5
4
∆H = + 66.4 kJ
O2(g) → NO(g) +
O2(g) → NO(g) +
3
2
3
2
H2O(g) + 226 kJ
H2O(g)
∆H = – 226 kJ
!
!
62.  CO2(g) + H2(g) + 41.2 kJ → CO(g) + H2O(g)
!
! + H O(g)
CO2(g) + H2(g) → CO(g)
2
∆H = + 41.2 kJ
64.  0.599 g H2
66.  23.9 kJ evolved
68.  9.74 kJ absorbed
78. 719 kg SiC
80. a) 50% from Pb and 50% from PbO2; (b) 11.7 g PbO2
82. 0.86 ton Cl2 (act)
84. All the silver from AgCl is recovered. 2.0 × 101 g additional AgCl could have been treated.
86. 19.1 lb C3H5(NO3)3
CHAPTER 11
 denotes problems assignable in OWL
2. A discrete line spectrum is one that consists of discrete, or separate, lines of color. Spectra of white light
do not have discrete lines. Rather, the colors blend into each other to form a continuous spectrum. A
rainbow is an example of a spectrum that does not have discrete lines. Discrete line spectra are not
normally encountered outside of chemistry and physics laboratories.
4. Both are forms of electromagnetic radiation.
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6.  (a) and (b) are quantized.
8.  Energy is absorbed; the electron moves further from the nucleus.
10. Discrete line spectra are evidence of quantized electron energy levels. Each discrete line represents one
of the possible energy transitions for an electron.
12. An atom is in its ground state when all electrons are at their lowest possible energies.
14. See Figure 11.7.
16. Hydrogen is the only atom that fits the Bohr model. Additionally, the Bohr electron should lose energy
and crash into the nucleus.
18. The principal energy levels of an atom are the main electron energy levels.
20.  Four sublevels: s, p, d, and f.
22. An orbital is a mathematically described region in space within an atom in which there is a high
probability that an electron will be found. An s orbital is spherical; a p orbital is dumbbell shaped. See
Figure 11.11 for sketches of shapes.
24.  1 s orbital, 3 p orbitals, and 5 d orbitals, for a total of 9 orbitals.
26. An orbital is a region in space where there is a high probability of finding an electron. This implies
correctly that there is a low but real probability of finding the electron outside that region—outside the ball.
28. The Pauli exclusion principle says, in effect, that no more than two electrons can occupy the same
orbital.
30. The quantum model of the atom gives no indication of the path of an electron. Furthermore, the orbital
is three dimensional, not two as suggested by a figure 8.
32.  There are seven 4f orbitals in an atom.
34. Energies of principal energy levels increase in the order 1, 2, 3,... Energies of sublevels increase in the
order s, p, d, f.
36. The models are consistent in identifying quantized energy levels. The quantum model substitutes
orbitals (regions in space) for orbits (electron paths) in the Bohr model. The quantum model goes beyond
the Bohr model in identifying sublevels.
38.  The symbol 3p4 means there are four electrons in the 3p sublevel.
40.  a) Ni is in group 8B/10 and period 4.
b) Mg is in group 2A/2 and period 3.
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42. The neon core, 1s22s22p6.
44.  a) titanium, b) neon
46. a) carbon, b) chlorine, c) arsenic
48.  Br: 1s22s22p63s23p64s23d104p5; Mn: 1s22s22p63s23p64s23d5
50. Ca: 1s22s22p63s23p64s2; Cu: 1s22s22p63s23p64s13d10
52.  a) Si: 1s22s22p63s23p2; b) P: [Ne] 3s23p3
54.  a) Sc: 1s22s22p63s23p64s23d1; b) Fe: [Ar] 4s23d6
56. Valence electrons are the electrons in the highest-energy incompletely-filled principal energy levels of
an atom.
58.  a) beryllium; b) selenium
60.  a) Carbon is in group 4A/14, period 2; b) Bromine is in group 7A/17, period 4
62.  Radon < xenon < neon < helium
64.  aluminum < silicon < phosphorus < chlorine
66.  a) One of: F, Cl, Br, I, At; b) one of: Li, Na, K, Rb, Cs, Fr; c) one of: He, Ne, Ar, Kr, Xe, Rn;
d) one of: Be, Mg, Ca, Sr, Ba, Ra
68.  a) Cesium; b) astatine; c) magnesium; d) argon
70.  (a), (c), (d)
72.  (c), (e)
74.  boron < aluminum < gallium < indium
76.  chlorine < sulfur < magnesium < sodium
78.  nitrogen < phosphorus < arsenic < antimony
80.  phosphorus < silicon < aluminum < sodium
82. The 3s1 electron in a sodium atom is in the third principal energy level. Removal of that electron to
form an ion leaves the second principal energy level as the highest occupied level. Therefore, we would
expect the ion to be smaller. The prediction is correct; the radius of the ion is 0.102 nm compared to 0.186
nm for the atom.
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84. Generally, an element is known as a metal if it can lose one or more electrons and become a positively
charged ion.
86. (a) Q and M, (b) D and E
88. J, W, and L
90. E > D > G
96. I: [Kr]5s24d105p5; W: [Xe]6s24f145d4 or [Xe]6s14f145d5
97. The first negatively charged electron is removed from a neutral atom, and the second electron is
removed from a positively charged ion.
98. The number of protons identifies an element. A neutral atom can gain or lose electrons, forming ions of
that element.
100. We expect magnesium to have a greater ionization energy than sodium and calcium because of the
trends in ionization energy explained in the text. Aluminum has a greater nuclear charge than magnesium,
but the additional electron is by itself in a 3p orbital, which makes it easier to remove.
101. (a) More metallic. Example: In Group 4A/14, carbon is a nonmetal, silicon and germanium are
metalloids, and tin and lead are metals. (b) Less metallic. Example: In Period 3, sodium, magnesium, and
aluminum are metals, silicon is a metalloid, and phosphorus, sulfur, chlorine, and argon are nonmetals.
CHAPTER 12
 denotes problems assignable in OWL
2.  a) anion; b) argon; c) Cl–
4.  S2–, Ar, K+, Ca2+
6.  Te2–, I–, Cs+, Ba2+
8. Your Lewis symbol for sulfur should show 6 dots, 2 each on 2 sides, and 1 each on 2 sides. The Lewis
symbol for sodium should show 1 dot. You will need 2 sodium atoms. Each of the sodium atoms will
transfer their electron to the sulfur atom. As a result, each sodium ion formed will have no valence
electrons, and the sulfide ion will have eight.
10.  3–2–2–3
12. An electron cloud is the area of space around an atom or between bonded atoms that is occupied by
electrons.
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14.  4; GeH4
16.  2; H2Te
18. Decrease
20. In a polar bond, the bonding electrons spend more time nearer the atom having the higher
electronegativity. This unequal sharing of electrons makes the bond polar. When two atoms of the same
element are bonded, the electronegativity difference between them is zero, and the bond is completely
nonpolar.
22.  Ge → Se (the arrowhead points toward the negative pole), Br ← Se, Br ← Ge. The Br ← Ge bond is
expected to be the most polar.
24.  Te → Se (the arrowhead points toward the negative pole), O ← Te, O ← Se. The O ← Te bond is
expected to be the most polar.
26.  Si < P < S < Cl
28. Multiple bond is a general term that includes both double and triple bonds. In a single bond, one
electron pair is shared between bonded atoms. In double and triple bonds, two and three electron pairs,
respectively, are shared between bonded atoms.
30. When a central atom has four electron pairs surrounding it, it has eight total electrons, which is an octet.
The four pairs can bond to a maximum of four atoms. If an atom in a molecule is truly “central,” it must be
bonded to a minimum of two other atoms.
32. 4 single bonds; 3 single bonds and 1 unshared pair; 2 single bonds and 2 unshared pairs; 2 double
bonds; 1 double bond and 2 single bonds; 1 double bond, 1 single bond, and 1 unshared pair; 1 triple bond
and 1 single bond.
34. In order to conform to the octet rule, each atom in a molecule must be surrounded by eight electrons.
Eight is an even number, and no combination of eights can result in an odd number.
36. A metallic bond is more similar to a covalent bond because, in both types of bonds, electrons are
shared. In an ionic bond, electrons are transferred.
38. Your sketch should show spheres to represent 1+ ions arranged in a repeating pattern. Among these
ions will be and equal number of 1– electrons that make up the electron sea.
40. An alloy is a solid mixture of two or more elements that has macroscopic metallic properties. Steel,
brass, and bronze are common alloys.
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44. (a) The electrostatic attractions between positive and negative ions results in a minimization of energy
in an ionic bond. (b) The stability of a noble gas electron configuration results in a minimization of energy
in the formation of a covalent bond.
46. A bond between sodium and sulfur is most likely to be ionic because it is between a metal and a
nonmetal. The bonds between fluorine and chlorine and between oxygen and sulfur are likely to be
covalent because they are between two nonmetals.
53. Rb2Se
55. According to the octet rule, a atom achieves a noble gas electron configuration in forming a bond.
There is no noble gas electron configuration in metallic bonding, which consists of positive ions in a sea of
electrons.
56. In either case, ionic or covalent bonding, the total energy of the system is minimized by the formation
of the bond. Atoms have noble gas electron configurations when these bonds are formed.
58. S—O; Al—O; Ca—O or Na—O; K—O. The group-to-group spread predicts Na—O is more polar than
Ca—O, but the period-to-period spread predicts Ca—O is more polar than Na—O.
60. Five or six electron pairs may surround a central atom if d orbitals are involved in bonding.
CHAPTER 13
 denotes problems assignable in OWL
F
2.
H
F
N
F
F
F
+
H
H
F
O
N
H
H
P
F
H
N
H
H
6. 
F
H
4. 
H
H
H
C
C
C
H
H
H
H
H
H
H
H
C
C
C
H
103
H
H
Instructor’s Manual to Accompany Cracolice/Peters 3/e
H
8. 
H
H
H
C
C
O
H
H
H
H
H
C
C
H
H
C
C
C
H
H
C
C
H
H
H
H
C
C
H
H
Br
F
F
Br
or
H
H
H
Br
H
H
Br
or
H
C
Br
10.
H
O
H
Br
H
H
H
H
F
C
C
C
H
H
Br
Br
H
H
F
H
C
C
C
H
H
Br
Br
or
H
F
H
H
C
C
C
H
H
Br
Br
or
H
F
Br
H
C
C
C
H
Br
H
H
or
F
Br
Br
C
C
C
H
H
H
or
H
H
H
Br
Br
C
C
C
H
F
H
H
H
or
or
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H
Br
Br
C
C
C
H
H
F
Br
H
Br
C
C
C
H
F
H
H
H
H
C
C
C
H
H
H
H
H
H
12.
H
H
H
or
Br
H
Br
C
C
C
H
H
F
H
H
H
C
C
C
H
or
H
H
H
H
H
H
H
H
H
C
C
C
O
H
H
or
C
H
C
H
H
H
H
H
H
H
C
H
C
C
C
H
O
H
H
H
C
C
H
O
H
H
H
C
C
C
H
or
or
H
H
or
H
O
C
H
H
H
14.
H
H
H
H
H
C
C
C
C
C
H
H
H
H
H
H
105
O
H
H
H
C
C
C
C
H
H
H
H
H
or
Instructor’s Manual to Accompany Cracolice/Peters 3/e
H
H
H
C
C
C
C
H
O
H
H
H
H
H
or
H
O
C
C
C
H
H
H
H
H
O
H
H
C
C
C
C
H
H
H
H
C
C
H
H
O
C
C
H
H
H
H
C
C
C
C
H
H
H
H
H
C
C
C
C
H
H
H
H
O
H
C
C
C
C
H
H
H
H
or
H
H
O
H
H
H
O
H
H
C
H
H
H
H
or
H
or
H
H
or
H
C
C
C
H
H
H
H
O
C
H
H
H
H
or
H
H
H
H
H
C
C
H
H
H
C
C
O
C
C
C
C
H
H
H
H
O
H
H
C
C
C
C
H
H
H
H
H
H
106
O
C
C
C
H
H
H
H
or
H
or
H
C
or
H
H
or
H
H
or
Instructor’s Manual to Accompany Cracolice/Peters 3/e
H
H
C
C
H
H
C
C
H
H
H
C
C
or
H
H
C
C
H
H
H
or
H
H
H
H
H
C
C
C
C
H
H
H
C
C
H
C
C
H
H
or
H
or
H
H
H
H
C
H
C
H
H
C
C
H
C
C
H
or
C
C
H
H
H
H
H
C
C
or
H
H
H
H
C
H
C
C
C
H
H
H
O
C
C
C
H
H
H
or
O
H
C
H
H
16.
H
–
O
18.
H
H
O
C
O
H
H
H
H
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Substance
Lewis
Diagram
20.  BF4–
–
F
F
B
ElectronPair
Geometry
Tetrahedral
Molecular
Geometry
Tetrahedral
Tetrahedral
Tetrahedral
Trigonal
Pyramidal
Tetrahedral
Trigonal
Pyramidal
Tetrahedral
Tetrahedral
Trigonal
planar
Trigonal
planar
Tetrahedral
Tetrahedral
Tetrahedral
F
F
CCl4
Cl
Cl
C
Cl
Cl
22.  PCl3
Cl
P
Cl
Cl
SeOF2
O
Se
F
F
24.  Each
C in
CH3CH2 COOH
H
26.  Each
C and O in
H
H
H
C
C
H
H
H
H
C
C
H
H
O
C
O
O
H
H
H
C
C
H
H
CH3CH2OCH2CH3
108
H
Bent
3D Ball-and-Stick
Representation
Instructor’s Manual to Accompany Cracolice/Peters 3/e
28.  Each
C and N in
H
H
O
C
C
Tetrahedral
N
Trigonal
pyramidal
H
CH3CONH2
H
30. C in OCCl2
Tetrahedral
H
O
Trigonal
planar
Trigonal
planar
Trigonal
planar
Trigonal
planar
C
Cl
Cl
A
B
32. Angular (Bent):
A
B
Bent (Angular):
B
A
34. Electron-pair: Linear; Molecular: Linear
36. Electron-pair: Tetrahedral; Molecular: Bent (Angular) or Electron-pair: Trigonal planar; Molecular
Angular (Bent)
38. Electron-pair: Tetrahedral; Molecular: Bent (Angular) or Electron-pair: Trigonal planar; Molecular
Angular (Bent)
40. A: 109°, 4 regions of electron density around C. B: 109°, 4 regions of electron density around O. C:
120°, 3 regions of electron density around C.
42. A: 120°, 3 regions of electron density around C. B: 109°, 4 regions of electron density around C. C:
109°, 4 regions of electron density around N.
44. a and c
F
B
46. 
F
F
a)
H
b)
N
H
trigonal planar
nonpolar
trigonal pyramidal
polar
H
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H
H
Ge
H
c)
48. 
H
F
a)
H
Be
F
tetrahedral
nonpolar
linear
nonpolar
bent
polar
trigonal planar
nonpolar
tetrahedral
nonpolar
bent
polar
trigonal planar
polar
O
H
b)
F
B
F
F
c)
Cl
Cl
50. 
Si
Cl
Cl
a)
H
Se
H
b)
O
C
c) H
H
52. e
54. Structure f is trigonal planar; c has a trigonal pyramidal shape.
56. The carbon atom can form four stable covalent bonds to other carbon atoms or to atoms of other
elements.
58. Hydro- in hydrocarbon refers to hydrogen. A hydrocarbon is a compound of hydrogen and carbon. The
ending -hydrate in carbohydrate refers to water. A carbohydrate is a compound of carbon and the elements
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that make up water, hydrogen and oxygen. One cannot be an example of the other. Carbohydrates contain
oxygen but hydrocarbons do not.
60. A Lewis diagram shows the structure of a molecule—the relative positions of the atoms to one
another—but it does not show the shape.
62. A line formula is one written as a “mini-Lewis diagram” in which the structure of the molecule is
suggested. Line formulas are especially useful for organic molecules because of the importance of their
structures, particularly for isomers.
64. An alcohol is an alkane with a hydroxyl group, —OH, substituted for a hydrogen. The properties of an
alcohol are the properties of the hydroxyl group.
66. Carboxylic acids contain a carboxyl group, —COOH. The geometry around the carbon is trigonal
planar, and it is bent around the oxygen.
O
C
O
H
73. Bond angles in C2H6 are all tetrahedral, which causes the molecule to be three-dimensional. The
geometry around both carbon atoms in C2H4 is trigonal planar, and the double bond between the carbon
atoms keeps all four hydrogen atoms in the same plane.
CHAPTER 14
 denotes problems assignable in OWL
2.  4.00 mol CO2 would have the largest volume.
4.  786 torr
6.  10.7 L
8.  1.32 mol
10.  73°C
12.  0.0297 mol
14.  17.3 L
16.  1.69 g/L
18.  32.2 g/mol
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20.  54.7 g/mol
22.  42.8 g/mol
24.  5.33 g/L
26.  26.6 L/mol
28.  47.8 L
30.  14.4 L CO2
32.  22.8 g C
34.  39.1 L Cl2
36.  17.4 L H2
38.  33.5 g Fe
40.  39.7 g Na
42.  9.60 L H2O
44.  12.6 L NO2
46.  28.4 L HCl
CHAPTER 15
 denotes problems assignable in OWL
2.  0.560 atm
4.  658 torr
6.  0.320 mol
8. Gas particles are very widely spaced; liquid particles are “touchingly close.” Density is mass per unit
volume. The same number of liquid particles will occupy a smaller volume than they occupy as a gas.
10. There is no space between water molecules; they cannot be forced closer together. Molecules in air are
widely separated; they can be pushed closer together.
12. The stronger the intermolecular attractions, the slower the evaporation rate, other things being equal.
Condensation rate therefore equals evaporation rate at a lower vapor concentration, which means lower
partial pressure at equilibrium.
14. Liquids with strong intermolecular attractions have high viscosity, an internal resistance to flow.
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16. Motor oil is more viscous than water. This predicts that intermolecular attractions are stronger in motor
oil, as strong attractions lead to internal resistance to flow, which is viscosity.
18. An isolated drop is normally spherical because of surface tension. When sitting on a plate, a drop tends
to be flattened by gravity. The honey drop remains closer to spherical than the water drop, indicating
stronger surface tension and stronger intermolecular attractions.
20. Soap reduces intermolecular attractions, thus lowering the surface tension of water. This makes the
soapy water able to penetrate fabrics and clean them throughout.
22. NO, N2O, NO2
24. Of the three compounds, only N2O is a liquid at –90°C. Therefore, only N2O possesses the property of
viscosity. If a solid is considered more “viscous” than a liquid, NO2 is the most viscous.
26. A compound with a large nonpolar section and an —OH or an —NH2 group at the end would exhibit
hydrogen bonding, but the induced dipole forces resulting from the large nonpolar section would dominate
the intermolecular attractive forces. An example is CH3CH2 CH2CH2 CH2CH2CH2OH.
28.  NOCl: dipole forces; NH2Cl: hydrogen bonds; SiCl4: induced dipole forces.
30. Ionic compounds have ionic bonds as the interparticle forces. Polar molecular compounds have dipole
forces acting between the particles. Ionic bonds are much stronger than any intermolecular forces,
including dipole forces. The ionic compound would have the higher melting point because more energy is
needed to overcome the stronger forces holding the solid together.
32. CCl4, because it is larger, as suggested by its higher molecular mass.
34. H2S, because it is slightly more polar.
36. For hydrogen bonding to occur, a hydrogen atom must be bonded to another element with a high
enough electronegativity to shift the bonding electron pair away from the hydrogen atom. Fluorine, oxygen,
and nitrogen are the only elements that satisfy this requirement.
38. a: induced dipole forces. b: dipole forces and induced dipole forces.
40. a, c, and d: dipole forces and induced dipole forces. b and e: induced dipole forces.
42.  d
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44. CO2 molecules are smaller than SO2 molecules, and CO2 molecules are linear and nonpolar while SO2
molecules are bent and polar. Both differences predict weaker intermolecular attractions and therefore
higher vapor pressure for CO2.
46. The term dynamic refers to the “active” character of the equilibrium. The rate of change from liquid to
vapor equals the rate of change from vapor to liquid. Although the concentrations remain constant, there is
continual change between the two states at the particulate level.
48.  Since C7H16 must be raised to a higher temperature than CH3COOCH3 to have a vapor pressure of
400 torr, the intermolecular forces are stronger in C7H16. The vapor pressure of CH3COOCH3 is 400 torr at
40°C. Since vapor pressure increases as temperature increases, its vapor pressure will be higher at 78°C.
50. Boiling point is the temperature at which the vapor pressure of a liquid is equal to the pressure above its
surface. Your vapor-pressure-versus-temperature curve should be similar to one from Figure 15.16. The
normal boiling point is found at a vapor pressure of 760 torr.
52. A gas can be condensed by increasing the pressure. Higher pressure forces the particles closer to one
another, where intermolecular attractive forces become significant, thus condensing a gas into a liquid.
54.  Since C8H18 must be raised to a higher temperature than CS2 to have a vapor pressure of 400 torr, the
intermolecular forces are stronger in C8H18. The normal boiling point of a liquid is the temperature at which
the vapor pressure of the liquid is 1 atm or 760 torr. Since C8 H18 must be raised to a higher temperature
than CS2 to have a vapor pressure of 760 torr, C8H18 will have the higher normal boiling point.
56. More energy is required to vaporize X, so it would have the higher boiling point and lower vapor
pressure.
58. Amorphous solids have no long-range ordering on the particulate level. Crystalline solids have particles
arranged in a repeating pattern. Polycrystalline solids have small orderly crystals arranged in a random
fashion. Because of the disorderly arrangement of particles in an amorphous solid, there is a range of
strengths of intermolecular attractive forces throughout the solid, and thus no definite physical properties.
Crystalline solids have distinct physical properties.
60. C, molecular; D, metallic
62.  2.20 kJ/g
64.  11.2 kJ
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66.  28.5 g
68.  – 1.22 kJ
70. 2.2 kJ
72. 151 J/g
74. 108 g Ag
76.  0.150 J/g • °C
78.  181 J
80.  – 35 J
82.  35°C
84. Vertically, temperature; horizontally, energy
86. E definitely; maybe D and F
88. G
90. The solid substance melts completely at constant temperature K.
92. O – N
94.  qtotal = 5.213 kJ + 9.333 kJ + 1.253 kJ = 15.799 kJ
96.  qtotal = 0.34 kJ + 10.39 kJ + 7.30 kJ = 18.03 kJ
98.  qtotal = (– 0.9709 kJ) + (– 1.199 kJ) + (– 0.6981 kJ) = – 2.868 kJ
CHAPTER 16
 denotes problems assignable in OWL
2. The properties of a solution will differ from those of the components. They will also vary depending on
the solution concentration.
4. Ions are present in a solution of an ionic compound; molecules are present in a solution of a molecular
compound unless the compound ionizes when it dissolves, as acids do. Then both ions and molecules are
present.
6. Generally, the substance present in the smallest amount is the solute. However, if gases or solids are
dissolved in a liquid, the liquid is usually called the solvent.
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8. If solute A is very soluble, its 10 grams per 100 grams of solvent concentration may be quite dilute
compared with the possible concentration. If solute B is only slightly soluble, its 5 grams per 100 grams of
solvent may be close to its maximum solubility, and therefore concentrated.
10. Drop a small amount of solute into the solution. If the solution is unsaturated, the solute will dissolve; if
saturated, it will simply settle to the bottom; if supersaturated, it will promote additional crystallization.
12. Any quantity units of solute divided by any quantity units of solvent may be used to express solubility.
14.  (a) supersaturated, (b) dilute, (c) miscible
16. A solute particle that attracts polar water molecules to itself is referred to as hydrated.
18. Attractive forces between solute and solvent particles promote dissolving. For an ionic solid solute in
water, the positively charged ions are attracted to the negatively charged regions of the polar water
molecules, and the negatively charged ions are attracted to the positively charged regions of the water
molecules.
20. On the macroscopic level, the process appears to be static; nothing seems to be happening. On the
particulate level, solute particles are continually changing between the solid and dissolved states, dissolving
and crystallizing. This constant change makes the process dynamic.
22. Solute ions pass from solute to solution faster than from solution to solute in an unsaturated solution.
24. Stirring decreases the rate of crystallization and thus decreases the time needed for formation of an
unsaturated solution. Any form of agitation, stirring or shaking, will be effective.
26. Increase the amount of surface area of the solid by dividing it into smaller pieces, agitate the solution to
decrease the rate of crystallization, or increase the temperature to speed particle movement.
28. (a) dimethyl ether and (d) hydrogen fluoride. Dimethyl ether is polar, as is water, and hydrogen fluoride
will match the hydrogen bonding found in water. Hexane and tetrachloroethane are nonpolar and more apt
to be soluble in nonpolar cyclohexane.
30. Water is polar and has hydrogen bonding. The same is true of ethanol, so it would probably not work. A
nonpolar solvent is more likely to dissolve what a polar solvent does not. Cyclohexane is more promising.
32.  (a) true, (b) false, (c) true
34.  12.2%
36.  44.2 g CuSO4
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38.  0.180 M AlBr3
40. 0.38 M Na2S2O3
42.  12.0 g ZnI2
44.  5.32 g Mn(C2H3O2)2
46.  176 mL
48. 0.29 L
50. 0.143 mol NaOH
52. 0.0141 mol H2SO4
54.  20.0%
56.  0.289 m
58.  12.9 g Al(NO3)3
60.  455 g H2O
62. The number of equivalents per mole depends on a specific reaction, not just the number of moles of H +
or OH– in a compound.
64.  a) 1 eq acid/mol; 1 eq base/mol
b) 2 eq acid/mol; 1 eq base/mol
66. 2 eq/mol Zn(OH)2; 1 eq/mol RbOH
68.  (a) 56.11 g KOH/eq; (b) 0.526 eq
70.  (a) 32.7 g H3PO4/eq; (b) 1.17 eq
72.  0.117 N
74. 0.234 N
76.  7.71 g Ba(OH)2
78.  (a) 0.1222 N; (b) 2.82 L
80. 1.96 eq
82. 0.104 L
84.  0.562 M
86.  82.0 mL
88.  0.165 M
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90. 0.94 N
92.  2.85 g Cu(OH)2
94.  401 mL
96. 0.336 L Cl2
98.  1.30 × 102 mL
100.  0.444 M KOH
102.  0.425 M NaOH
104.  0.109 M KOH
106. 0.444 N KOH
108. 0.425 N NaOH
110. 1.03 N acid
112.  0.290 N
114.  0.253 N KOH
116. 72.0 g/eq
118. A colligative property of a solution is independent of the identity of the solute. Specific gravity, with
opposite effects from different solutes, is not a colligative property.
120.  0.1522 m; Tb = 80.10°C + 0.385°C = 80.49°C
122.  0.6713 m; – 1.25°C
124.  0.2426 m; 0.824°C/m
126.  3.6 × 102 g/mol
128.  66.3 g/mol
130.  2.18°C/m
137. Distillation is one method. It is used to separate petroleum into its components, which include natural
gas, gasoline, lubricating oil, asphalt, and many other products.
144. 7.15 g Na2 CO3 • 10 H2O
147. 0.362 g Ni(OH)2 will precipitate
149. 0.366 M OH–
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CHAPTER 17
 denotes problems assignable in OWL
2.  a) HCN(aq): Brønsted-Lowry acid and proton source
H2O(l): Brønsted-Lowry base and proton remover
b) CH3NH2(aq): Brønsted-Lowry base and proton remover
H2O(l): Brønsted-Lowry acid and proton source
4.  a) HClO(aq) + H2O(l) → ClO–(aq) + H3O+(aq)
b) C6H15O3N(aq) + H2O(l) → C6H15O3NH+(aq) + OH–(aq)
6.  (a) CCl4: (4) Neither a Lewis acid nor a Lewis base
(b) F–: (2) Lewis base
(c) Cu2+: (1) Lewis acid
(d) Fe3+: (1) Lewis acid
(e) O2: (2) Lewis base
8. Aluminum in aluminum chloride is able to accept an electron pair, so it qualifies as a Lewis acid. The
chloride ion can contribute the electron pair to the bond, so it is a Lewis base.
–
Cl
Cl
–
Cl
Al
+
Cl
Cl
Cl
Al
Cl
Cl
10.  (a) CH3COO–; (b) HCN: (c) HS–; (d) H2 CO3
12.  HF(aq): Brønsted-Lowry acid; CN–(aq): Brønsted-Lowry base;
F–(aq): Brønsted-Lowry base; HCN(aq): Brønsted-Lowry acid.
The conjugate base of HF(aq) is F–(aq). The conjugate acid of CN–(aq) is HCN(aq).
14.  CH3COO–(aq): Brønsted-Lowry base; HS–(aq): Brønsted-Lowry acid;
CH3COOH(aq): Brønsted-Lowry acid; S2–(aq): Brønsted-Lowry base.
The conjugate acid of CH3 COO–(aq) is CH3COOH(aq). The conjugate base of HS–(aq) is S2–(aq).
16. HClO(aq) and ClO–(aq); CO32–(aq) and HCO3–(aq)
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18. H2PO4–(aq) and HPO42–(aq); HCO3–(aq) and H2CO3(aq)
20. A strong acid loses protons readily; a weak acid does not lose protons easily. Strong acids are at the top
of the left column in Table 17.1 and weak acids are at the bottom.
22.  (a) Sulfurous acid, H2SO3; (b) Hydrofluoric acid, HF
24.  (a) Dihydrogen phosphate ion, H2PO4–; (b) Cyanide ion, CN–
26.  HCN(aq) + CH3 COO–(aq)  CN–(aq) + CH3COOH(aq); Reverse
28.  F–(aq) + HCO3–(aq)  HF(aq) + CO32–(aq); Reverse
30. H3PO4(aq) + CN–(aq)  H2PO4–(aq) + HCN(aq); Forward
32. H2BO3–(aq) + NH4+(aq)  H3BO3(aq) + NH3(aq); Reverse
34. HPO42–(aq) + HC2H3O2(aq)  H2PO4–(aq) + C2H3O2–(aq); Forward
36. See the five items listed in Section 17.7.
38. Water ionizes very slightly and does not produce enough ions to light an ordinary conductivity device.
With a sufficiently sensitive detector, water displays a very weak conductivity.
40.  (a) [OH–] = 10–10 M; (b) Acidic
42.  (a) [H+] = 10–3 M; (b) Acidic
44. Strongly acidic, pH < 4; weakly acidic, 4 ≤ pH < 6; strongly basic, 10 ≤ pH; weakly basic, 8 ≤ pH < 10;
neutral or near neutral, 6 ≤ pH < 8
46. If pH = x, then [H+] = 10–x
pH
pOH
[H+]
[OH–]
Classification
48. 
8
6
10–8
10–6
Weakly basic
50. 
1
13
10–1
10–13
Strongly acidic
52. 
12
2
10–12
10–2
Strongly basic
54. 
4
10
10–4
10–10
Weakly acidic
56. 
6.62
7.38
2.4 × 10–7
4.2 × 10–8
58. 
3.04
10.96
9.1 × 10–4
1.1 × 10–11
60. 
8.46
5.54
3.5 × 10–9
2.9 × 10–6
62. 
1.14
12.86
7.2 × 10–2
1.4 × 10–13
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CHAPTER 18
 denotes problems assignable in OWL
2. Rates of change in opposite directions are equal in an equilibrium. Examples include the liquid-vapor
equilibrium in Section 15.4 and the solution equilibrium in Section 16.3.
4. Neither matter nor energy may enter or leave a closed system.
6. The system is not closed—water enters and leaves—so it is not an equilibrium.
8. Colliding particles must have sufficient energy and proper orientation to react.
10. ∆E is negative; the reaction is exothermic. ∆E = c – b. Activation energy = a – b.
12. The activation energy is greater for the reverse reaction: a – c.
14. See text for a discussion of activation energy. All other things being equal, the reaction with the lower
activation energy will be faster because a larger fraction of reacting particles will be able to engage in
reaction-producing collisions.
16. The statement refers to the fraction of all possible reacting particles that has enough energy to engage in
a reaction-producing collision. In Figure 18.4,
shaded area
number of particles with enough energy to react
=
total area
total number of particles
18. A catalyst is a substance that increases reaction rate by lowering activation energy. The catalyst is not
! changed in!the reaction.
permanently
20. Rate varies directly with reactant concentration. As A concentration increases, rate increases; as B
concentration decreases, rate decreases.
22. At the beginning, when both reactants are at highest concentration.
24. Nitrogen and hydrogen concentrations decrease; ammonia concentration increases.
26.  The reaction must run in the reverse direction to reestablish equilibrium. The concentration of PCl3
will increase.
28.  The reaction must run in the forward direction to reestablish equilibrium. The concentration of H2
will decrease.
30.  The reaction must run in the forward direction to reestablish equilibrium. The concentration of CO
will increase.
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32.  The reaction must run in the reverse direction to reestablish equilibrium. The number of moles of Br2
will decrease.
34.  The reaction must run in the reverse direction to reestablish equilibrium. The number of moles of O2
will increase.
36.  The reaction must run in the forward direction to reestablish equilibrium. The concentration of Cl2
will increase.
38.  The reaction must run in the forward direction to reestablish equilibrium. The concentration of I2 will
decrease.
40.  The production of CH2Cl2 is favored by: (a) increasing the temperature.
42.  K = [NH3] [H2S]
44.  K =
[SO 3 ]2
[SO 2 ]2 [O 2 ]
46.  K =
!
[H 3O + ] [F – ]
[HF]
48.  K =
!
[(CH 3 ) 3 NH + ] [OH – ]
[(CH 3 ) 3 N]
50.  K = [Mg2+] [OH–]2
! In a K expression, the species on the right are always in the numerator, and those on the left are in the
52.
denominator. As the example equation is written, K =
N2(g) + 3 H2(g), K =
[N 2 ][H 2 ] 3
[NH 3 ] 2
[NH 3 ] 2
[N 2 ][H 2 ] 3
. If the equation is reversed, 2 NH3(g) 
.
!
54.  (b) The reverse reaction is favored at equilibrium.
!
[H + ][C 2H 3O 2 " ]
56. HC2H3O2  H+ + C2H3O2−. The equilibrium constant, K =
, will be small.
[HC 2H 3O 2 ]
A weak acid ionizes only slightly, producing very small concentrations of the numerator species compared
!
with the almost unchanged concentration of the denominator
species.
58. a) Forward, with large K. HCl is a strong acid and ionizes almost completely. b) Reverse, with small
K. BaSO4 is a low-solubility solid and releases few ions.
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60.  6.74 × 10–9
62.  1.69 × 10–8
64. (a) s = 9.3 × 10−5 mol/L
(b)
9.3 " 10#4 g CaCO 3
100 mL
66.  s = 2.3 × 10−5 mol/L
68.  s = 1.9 × 10−7
!
70.  % ionization =2.9% ; Ka = 4.1 × 10−4
72.  pH = 3.89
74.  pH = 4.68
76. 
[HCN]
= 0.49
[CN – ]
78.  K = 2 × 101
! 82. Increase [AB], heat the system, and introduce catalyst.
CHAPTER 19
 denotes problems assignable in OWL
2. An electric current through a wire is a one-way movement of electrons. Electrolysis is a two-way
movement of charged ions in a solution.
4. No. An electrolytic cell must be connected to an outside source of electricity for electrolysis to occur. It
cannot function as a voltaic cell, which spontaneously produces an electric current when connected to an
outside circuit.
6.  Species oxidized: Mg(s); species reduced: Co2+(aq); electrons are transferred from Mg(s) to Co2+(aq).
8. a, b, c: oxidation; d: reduction
10. Oxidation
12. Reduction
14.  The oxidation half-reaction is Ag(s) → Ag+(aq) + e–.
The reduction half-reaction is Br2(l) + 2 e– → 2 Br–(aq).
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16. 
2×
[Fe2+(aq)
→
Fe3+(aq) + e–]
Oxidation
Cl2(g) + 2 e–
→
2 Cl–(aq)
Reduction
2 Fe2+(aq) + Cl2(g)
→
2 Fe3+(aq) + 2 Cl–(aq)
18.  +4, +6, –1
20.  +3, –1, 0
22. a) Copper reduced from +2 to 0. b) Cobalt reduced from +3 to +2.
24. a) Sulfur oxidized from +4 to +6. b) Phosphorus oxidized from –3 to 0.
26. a) Fluorine oxidized from –1 to 0. b) Manganese reduced from +6 to +4.
28.  Sulfur in SO42–: +6; nickel in Ni2+: +2; sulfur in SO2: +4; nickel in NiO2: +4.
Element oxidized: nickel; element reduced: sulfur; oxidizing agent: SO42–; reducing agent: Ni2+.
30.  Manganese in Mn2+: +2; nitrogen in HNO3: +5; manganese in MnO2: +4; nitrogen in NO: +2.
Element oxidized: manganese; element reduced: nitrogen; oxidizing agent: HNO3 ; reducing agent: Mn2+.
32.  Strongest oxidizing agent: F2; weakest oxidizing agent: Fe2+; strongest reducing agent: Fe;
weakest reducing agent: F–.
34.  Strongest oxidizing agent: Cl2; weakest oxidizing agent: Cd2+; strongest reducing agent: Cd;
weakest reducing agent: Cl–.
36.  Ni2+ + Cd  Ni + Cd2+; Forward
38.  Cl2 + Fe  2 Cl– + Fe2+; Forward
40.  2 H + + Mg  H2 + Mg2+; Forward
42. See the five items listed in Section 19.7.
44. 
46. 
2×
[Zn
→
Zn2+ + 2 e–]
Sb2O5 + 6 H+ + 4 e–
→
2 SbO+ + 3 H2O
2 Zn + Sb2O5 + 6 H+
→
2 Zn2+ + 2 SbO+ + 3 H2O
Ni2+ + 2 H2O
→
NiO2 + 4 H+ + 2 e–
Sn2+ + 2 e–
→
Sn
Ni2+ + 2 H2O + Sn2+
→
NiO2 + 4 H+ + Sn
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Instructor’s Manual to Accompany Cracolice/Peters 3/e
48. 
2×
50. 
52. 
Cl– + 3 H2O
→
ClO3– + 6 H+ + 6 e–
[NO3– + 4 H+ + 3 e–
→
NO + 2 H2O]
Cl– + 2 NO3– + 2 H +
→
ClO3– + 2 NO + H2O
SO42– + 4 H+ + 2 e–
→
SO2 + 2 H2O
Hg
→
Hg2+ + 2 e–
SO42– + 4 H+ + Hg
→
SO2 + 2 H2O + Hg2+
Pb2+ + 2 H2O
→
PbO2 + 4 H + + 2 e–
Zn2+ + 2 e–
→
Zn
Pb2+ + 2 H2O + Zn2+
→
PbO2 + 4 H + + Zn
CHAPTER 20
 denotes problems assignable in OWL
2.  (a)
0
–1 e ;
(b) gamma; (c) gamma is the most penetrating and alpha is the least penetrating
4.  b, d, and e
!6.  b, d, and e
8. Ionizing radiation refers to radioactive emissions energetic enough to cause particles to ionize. Not all
radioactive emissions have sufficient energy to cause the target particles to lose an electron.
10. Radiation can be detected by its ability to expose photographic film, ionize a gas, or cause a substance
to emit light.
12. A scintillation counter is a device used to measure radiation. Particles in a transparent solid absorb
energy from the radiation, and subsequently release some of the energy in the form of light. This light is
detected and measured by the counter.
14. A gamma camera is stationary, taking a single image, and a scanner moves, taking multiple images of
the area to be observed. Their principal advantage is that they allow medical personnel to “see” inside the
body without surgery.
16. Background radiation is potentially dangerous, but most of it is unavoidable. You can avoid voluntary
exposure to radiation primarily by avoiding tobacco smoke.
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